differential equations?

Question

how do I determine whether y=5tan(5x) is a solution of y'= 25+y^2

Answer 1

just check weather it is a solution by substitute the value of y which is y=5tan(5x)--->(1) abd its y' which is y'=25sec^2(5x)-->(2)

show:
y'=25+y^2
substitute: (1) & (2)

25sec^2(5X)?=25+25(tan^2(5x))

25sec^2(5X)?=25(1+tan^2(5x))

where 1+tan^2(5x)=sec^2(5x) -- a trigonometric identity

25sec^2(5X)=25sec^2(5X)

we prove now that is a solution

Answer 2

Compute both sides and see whether or not they are equal. :)

For the LHS, use the Chain Rule. For the RHS, use trigonometric identities as needed.